Questions
A tuning fork of frequency 392 Hz, resonates with 50 cm length of a string under tension (T). If length of the string is decreased by 2%, keeping the tension constant, the number of beats heard when the string and the tuning fork made to vibrate simultaneously is
detailed solution
Correct option is C
n∝1l⇒Δnn=−ΔllIf length is decreased by 2% then frequency increases by 2% i.e., n2−n1n1=2100⇒n2−n1=2100×n1=2100×392=7.8≈8.Talk to our academic expert!
Similar Questions
A metal wire of diameter 1.5 mm is held on two knife edges separated by a distance 50 cm the tension in the wire is 100 N the wire vibrating with its fundamental frequency and vibrating tuning fork together produces 5 beats per second. The tension in the wire is then reduced to 81 N, when the two are excited, beats are heard at the same rate. Calculate frequency of the fork.
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